1 - Since electrοns can be transferred frοm οur hair tο the ballοοn , electrοns cannοt be transferred frοm ballοοn tο οur hair because. This is an illustratiοn οf charging by cοnductiοn.
2 - Since the rubber οn the ballοοn is significantly less cοnductive than the hair, electrοns will nοt easily escape the ballοοn because οf this.
3 - Neutrοns are electrically neutral , neutrοns dοesn't participate in this prοcess.
What is charging by cοnductiοn?A charged οbject must cοme intο cοntact with a neutral οbject tο cοnduct electricity. As a result, when twο charged cοnductοrs cοme intο cοntact, the charge is split between the twο cοnductοrs, charging the uncharged cοnductοr.
When twο neutral οbjects are rubbed against οne anοther, electrοns are transferred. The οbject that has a strοnger affinity fοr electrοns will take electrοns frοm the οther οbject, and the twο becοme charged in οppοsitiοn. In this instance, the electrοns frοm the hair are taken up by the ballοοn , which nοw has an excess οf electrοns and a negative charge cοmpared tο the hair's current electrοn shοrtage and pοsitive charge.
2- Since the rubber οn the ballοοn is significantly less cοnductive than the hair, electrοns will nοt easily escape the ballοοn because οf this.
3- Neutrοns are electrically neutral , neutrοns dοesn't participate in this prοcess.
4-Insulating materials may becοme electrically charged when they cοme intο cοntact with οne anοther. Negatively charged electrοns can "rub οff" οne material and "rub οn" tο anοther. After bοth things have the same quantity οf οppοsite charges, the substance that gets electrοns becοmes negatively charged, and the material that lοses electrοns becοmes pοsitively charged.
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how to find the minimum thickness of a film such that reflected light undergo constructive interference
The minimum thickness of the film for constructive interference of reflected light would be t = 3*600/(2*1.4) = 850 nm.
The minimum thickness of a film required for constructive interference of reflected light can be calculated using the formula t = m*λ/(2*n),
where t is the minimum thickness of the film, m is the order of interference, λ is the wavelength of the light, and n is the index of refraction of the film.
For example, if the order of interference is 3, the wavelength of the light is 600 nm, and the index of refraction is 1.4,
the minimum thickness of the film for constructive interference of reflected light would be t = 3*600/(2*1.4) = 850 nm.
Constructive interference of reflected light occurs when the phase difference between the two waves is equal to an integral multiple of 2π.
This can be determined using the formula Δφ = (2π*m)/(λ*n), where Δφ is the phase difference, m is the order of interference, λ is the wavelength of the light, and n is the index of refraction of the film.
To achieve constructive interference, the minimum thickness of the film can be determined by ensuring that the phase difference is equal to an integral multiple of 2π.
The minimum thickness of a film required for constructive interference of reflected light can be calculated using the formula t = m*λ/(2*n),
where t is the minimum thickness of the film, m is the order of interference, λ is the wavelength of the light, and n is the index of refraction of the film.
Constructive interference can be achieved by ensuring that the phase difference between the two waves is equal to an integral multiple of 2π.
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An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?
Angular Momentum Kinetic Energy
(A) Increases Increases
(B) Increases Remains Constant
(C) Remains Constant Increases
(D) Remains Constant Remains Constant
(E) Decreases Remains Constant
An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled closer to the body, the angular momentum of the skater will remain constant while the kinetic energy of the skater increases. The correct option is C.
The angular momentum of the skater is given by
[tex]L = I\omega[/tex],
where I is the moment of inertia of the skater and ω is the angular velocity.
When the skater pulls their arms in, their moment of inertia decreases due to the decreased distance between their body and the axis of rotation.
According to the conservation of angular momentum, the product of the moment of inertia and angular velocity must remain constant. Therefore, if the moment of inertia decreases, the angular velocity must increase to keep the angular momentum constant.
The kinetic energy of the skater is given by
[tex]K = (1/2)I\omega^2[/tex]
As the moment of inertia decreases and the angular velocity increases, the kinetic energy of the skater also increases because it is proportional to the square of the angular velocity.
Therefore, the correct answer is: (C) Remains Constant Increases. The angular momentum remains constant, while the kinetic energy increases due to the increased angular velocity.
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pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?
The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.
The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.
A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.
In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.
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a 10.0-mf capacitor is fully charged across a 12.0-v bat- tery. the capacitor is then disconnected from the battery and connected across an initially uncharged capacitor with capacitance c. the resulting voltage across each capacitor is 3.00 v. what is the value of c?
The value of uncharged capacitor in series with a 10.0-microfarad capacitor, given that each capacitor has a voltage of 3.00 volts, can be calculated using the formula for equivalent capacitance in series and formula for charge on a capacitor. The value of c is approximately 4.00 microfarads.
To determine the value of c, which is [tex]1/Ceq = 1/C1 + 1/C2[/tex] . Initially, the 10.0-microfarad capacitor has a charge of [tex]Q = CV = (10.0 * 10^{-6 }F) * 12.0 V = 1.20 * 10^{-4} C[/tex].
When it is connected in series with uncharged capacitor with capacitance c, charge is shared between the two capacitors. The charge on 10.0-microfarad capacitor is also equal to the charge on uncharged capacitor, which is given by [tex]Q = (3.00 V) * C[/tex] .
Equating the two expressions for Q and solving for c, we get [tex]C = Q/3.00[/tex] [tex]V = (1.20 * 10^{-4 C}) / (3.00 V) = 4.00 * 10^{-5 F}[/tex]. Therefore, value of c is approximately 4.00 microfarads.
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two stationary point charges q1 and q2 are shown in the figure along with a sketch of some field linesrepresenting the electric field produced by them. what can you deduce from the sketch?
From the sketch, we can deduce that the two charges q1 and q2 are of opposite signs, as field lines start at the positive charge q1 and end at the negative charge q2. The field lines also indicate that the magnitude of the electric field produced by q1 is larger than that of q2.
Additionally, the field lines show that the electric field lines near the charges are denser, indicating a stronger electric field intensity near the charges. The direction of the electric field points from q1 to q2, which is consistent with the direction of the force that a positive test charge would experience if placed in the field. The field lines also show that the electric field is radial, i.e., the field lines point directly away from or towards each charge in a straight line, which is a characteristic of the electric field produced by a point charge. Finally, the density of the field lines decreases with distance from the charges, indicating that the electric field strength decreases with distance from the charges, following an inverse-square law.Learn more about electric field at: https://brainly.com/question/14372859
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a system releases 690 kj of heat and does 110 kj of work on the surroundings. part a what is the change in internal energy of the system?
A system releases 690 kj of heat and does 110 kj of work on the surroundings then part a what i the change in internal energy of the system -800 kJ.
The change in internal energy of the system can be calculated using the formula
ΔU = Q - W,
where ΔU is the change in internal energy, Q is the heat exchanged, and W is the work done.
In this case, the system releases 690 kJ of heat (Q = -690 kJ) and does 110 kJ of work on the surroundings (W = 110 kJ).
So, ΔU = -690 kJ - 110 kJ = -800 kJ.
The change in internal energy of the system is -800 kJ.
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a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. the coefficients of static and kinetic friction are 0.6 and 0.4, respectively. the worker pushes with a force of 600 n. the frictional force exerted by the surface is
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface on the crate is 588 N.
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface can be calculated as follows:
The weight of the crate = m × g = 100 kg × 9.8 m/s² = 980 N
Force applied by the worker = F = 600 N
The force of friction acting on the crate is given by the following formula:
Ff = μF
Where, μ is the coefficient of friction, F is the normal force acting on the crate.
Notes: The normal force is equal and opposite to the weight of the crate. i.e., N = 980 N1. The frictional force exerted by the surface on the crate is the static frictional force initially. Hence, we use the coefficient of static friction for our calculation.
2. If the force applied by the worker is not enough to overcome the static frictional force, then the crate will not move and the frictional force will remain static friction.
3. Once the crate starts moving, the static friction will convert to kinetic friction. Hence, we will use the coefficient of kinetic friction if the force applied by the worker is greater than the force of static friction. Initially, the force applied by the worker is less than the force of static friction, hence the frictional force exerted on the crate will be the static frictional force.
Frictional force = Ff = μN
The normal force acting on the crate = Weight of the crate = 980 N
Frictional force =
Ff = μN
= 0.6 × 980 N
= 588 N
Therefore, the frictional force exerted by the surface on the crate is 588 N.
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To determine the location of her center of mass, a physics student lies on a lightweight plank supported by two scales 2.50m apart, as indicated in the figure . If the left scale reads 290 N, and the right scale reads 112 N. What is the student's mass and find the distance from the student's head to her center of mass.
The location of her centre of mass, a physics student lies on a lightweight plank supported by two scales 2.50m apart, as indicated in the figure. If the left scale reads 290 N and the right scale reads 112 N The student's mass is approximately 41 kg, and the distance from her head to her centre of mass is approximately 0.696 m.
To determine the student's mass, we can sum up the readings from both scales, which are measures of force (Newtons) and then convert it to mass using the gravitational acceleration (g = 9.81 m/s²).
Step 1: Calculate the total force acting on the plank:
Total Force = Force_left_scale + Force_right_scale
Total Force = 290 N + 112 N
Total Force = 402 N
Step 2: Convert the total force to mass using gravitational acceleration:
Mass = Total Force / g
Mass = 402 N / 9.81 m/s²
Mass ≈ 41 kg
Now, to find the distance from the student's head to her centre of mass, we'll use the principle of torque equilibrium.
Step 3: Set up the torque equation:
Torque_left_scale = Torque_right_scale
Force_left_scale × Distance_left_scale = Force_right_scale × Distance_right_scale
Let x be the distance from the student's head to her centre of mass. Then, the distance from the left scale to the centre of mass is x, and the distance from the right scale to the centre of mass is (2.50 - x).
Step 4: Plug in the known values and solve for x:
290 N × x = 112 N × (2.50 - x)
Step 5: Simplify the equation and solve for x:
290x = 112(2.50) - 112x
290x + 112x = 112(2.50)
402x = 280
x ≈ 0.696 m
The student's mass is approximately 41 kg, and the distance from her head to her centre of mass is approximately 0.696 m.
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logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. what is the average density of a uniform-diameter log that floats with 20.0% of its length above water?
Uneven-diameter logs that float with 20.0% of their length above water have an average density of 0.8g/cm3. The density is the proportion of weight to capacity.
An item it's far less compact that liquid may be supported up liquid water, and hence it floats. More dense objects can sink when submerged in water. Less dense logs float whereas more thick logs sink. A body can change its condition of rest or motion by the application of force
Instead of obliquely reading from either the side, read the scale stick straight from of the end of both the log. → The diameter of a log is only ever calculated within the bark. Employ a log measuring rod to determine the log's small end's "diameter from within bark," also known as "d.i.b."
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a mass-spring oscillating system undergoes shm with a period t. what is the period of the system if the amplitude is doubled?
The period of a mass-spring oscillating system undergoing SHM with a period t, when the amplitude is doubled, is still t.
The period of a mass-spring oscillating system undergoing simple harmonic motion (SHM) is determined by the spring constant and mass of the system.
When the amplitude of the system is doubled, the period of the system remains the same, regardless of the amplitude. This means that the period of a mass-spring oscillating system undergoing SHM with a period t, when the amplitude is doubled, is still t.
To understand why the period remains the same, consider the equation for simple harmonic motion:
x(t) = A cos (2πft).
This equation describes the displacement of an object over time and is based on the principle that any system undergoing SHM oscillates about a fixed point at a constant frequency.
The frequency of the system is inversely proportional to the period, and is determined by the spring constant and mass of the system.
Increasing the amplitude of the system does not affect the frequency or period of the oscillations.
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the intensity of the sound of a television commercial is 10 times greater than the intensity of the television program it follows. by how many decibels does the loudness increase?
The television commercial loudness increases by 10 decibels.
Increase in the Intensity of soundThe decibel (dB) scale is a logarithmic measure of sound intensity. The intensity of a sound is measured in watts per square meter and the decibel scale is a way to express the relative loudness of a sound, compared to a reference level.
A 10 dB increase in intensity is a 10-fold increase in sound power. This means that a sound with an intensity of 10 watts per square meter is 10 times louder than a sound with an intensity of 1 watt per square meter.
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water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. the density of water is 1 000 kg/m3. determine its average velocity. multiple choice question. 20 m/s 200 m/s 0.02 m/s 2 m/s 0.2 m/s
Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.
According to the question:
cross-sectional area of the pipe = 0.002m²
Mass flowrate = 4 kg/s
Density of water = 1000 kg/m³
We are asked to find, average velocity =?
Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.
As a result, the water's average flow rate through the pipe is provided by:
v = m / (ρ × A)
where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:
v = 4 / (1000 × 0.002)
v = 2m/s
Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.
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Correct question is:
Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.
20 m/s
200 m/s
0.02 m/s
2 m/s
0.2 m/s
Why is momentum not conserved in real life situations
Momentum is not always conserved in real-life situations because external forces can act on a system and change its momentum.
For example, when two cars collide, friction and air resistance can cause the momentum of the system to change. Similarly, when a ball is thrown in the air, gravity and air resistance act on it and cause its momentum to change. Other factors such as deformation, energy loss, and imperfect collisions can also cause momentum to be lost or gained. Therefore, while momentum is a useful concept in physics, it is important to consider the impact of external factors when analyzing real-world situations.
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